If `z=(1+i)/sqrt2` , then the value of `z^1929` is

Options

A. 1+i
B. `-1`
C. `(1+i)/2`
D. `(1+i)/sqrt2`

Answer

To solve the problem, we need to find the value of \( z^{1929} \) where \( z = \frac{1+i}{\sqrt{2}} \).

### Step-by-Step Solution:

1. **Express \( z \) in terms of its components**:
\[ z = \frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}} \]

2. **Recognize the trigonometric form**:
We can express \( \frac{1}{\sqrt{2}} \) and \( \frac{i}{\sqrt{2}} \) in terms of cosine and sine. We know:
\[ \frac{1}{\sqrt{2}} = \cos\left(\frac{\pi}{4}\right) \quad \text{and} \quad \frac{1}{\sqrt{2}} = \sin\left(\frac{\pi}{4}\right) \]
Thus, we can rewrite \( z \) as:
\[ z = \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) \]

3. **Convert to exponential form**:
Using Euler's formula, we can express \( z \) as:
\[ z = e^{i \frac{\pi}{4}} \]

4. **Raise \( z \) to the power of 1929**:
\[ z^{1929} = \left(e^{i \frac{\pi}{4}}\right)^{1929} = e^{i \frac{1929 \pi}{4}} \]

5. **Simplify the exponent**:
To simplify \( \frac{1929 \pi}{4} \), we divide 1929 by 4:
\[ 1929 \div 4 = 482 \quad \text{remainder } 1 \]
Thus,
\[ 1929 = 4 \times 482 + 1 \]
Therefore,
\[ \frac{1929 \pi}{4} = 482 \pi + \frac{\pi}{4} \]

6. **Use periodicity of sine and cosine**:
Since \( e^{i \theta} \) has a period of \( 2\pi \), we can reduce \( 482 \pi \):
\[ e^{i(482 \pi + \frac{\pi}{4})} = e^{i \frac{\pi}{4}} \quad (\text{because } 482 \text{ is even, } e^{i 2k\pi} = 1) \]

7. **Final result**:
Thus, we have:
\[ z^{1929} = e^{i \frac{\pi}{4}} = \cos\left(\frac{\pi}{4}\right) + i \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} = z \]

### Conclusion:
The value of \( z^{1929} \) is:
\[ \frac{1+i}{\sqrt{2}} \]